Summary: Although most of our prior experiments with recursion have emphasized recursion over lists, it is also possible to use other values as the basis of recursion. In this laboratory, you will explore the use of natura l numbers (non-negative integers) as the basis of recursion.
a. Make sure you understand the first reading on recursion and the second reading on recursion.
b. Start DrScheme.
Using recursion over natural numbers, define and test a recursive Scheme
(power-of-two power) that takes a natural
number (an integer greater than or equal to 0)
as its argument and returns the result of raising 2 to the power of
that number. For example,
> (power-of-two 3) 8 > (power-of-two 10) 1024 > (power-of-two 1) 2
It is possible to define this procedure non-recursively, using Scheme's
expt procedure, but the point of the exercise is to
Define and test a Scheme procedure,
val), that takes a natural number as argument and returns
a list of all the natural numbers less than or equal to that number, in
> (count-down 5) (5 4 3 2 1 0) > (count-down 0) (0)
Define and test a Scheme procedure,
count), that takes two arguments, the second of which is a
natural number, and returns a list consisting of the specified number of
repetitions of the first argument:
> (fill-list 'sample 5) (sample sample sample sample sample) > (fill-list (list 'left 'right) 3) ((left right) (left right) (left right)) > (fill-list null 1) (()) > (fill-list null 2) (() ())
Define and test a recursive Scheme procedure that takes a natural number
as argument and returns a list of all the natural numbers that are
strictly less than the argument, in ascending order. (The traditional
name for this procedure is
iota -- another Greek letter.)
> (iota 3) (0 1 2) > (iota 5) (0 1 2 3 4) > (iota 1) (0)
You may recall the
count-from procedure from
the reading on recursion
over natural numbers. That procedure is also reproduced
at the end of this lab.
What is the value of the call
(count-from -10 10)?
a. Write down what you think that it should be.
b. Copy the definition of
into DrScheme and use it to find out what the call
count-from, define and test a Scheme procedure that
takes a natural number as argument and returns a list of all the natural
numbers that are strictly less than the argument, in ascending order.
Note that your procedure must use
count-from as a helper.
Here is the definition of a procedure that computes the number of digits in
the decimal representation of
(define number-of-decimal-digits (lambda (number) (if (< number 10) 1 (+ (number-of-decimal-digits (quotient number 10)) 1))))
a. Test this procedure.
The definition of
number-of-decimal-digits uses direct
b. Describe the base case of this recursion.
c. Identify and describe the way in which a simpler instance of the problem is created for the recursive call. That is, explain what problem is solved recursively and why you know that that problem is simpler.
d. Explain how the procedure correctly determines that the decimal numeral for the number 2000 contains four digits.
e. What preconditions does
on its argument?
1. Finish your work on the previous lab.
power-of-two to permit negative exponents.
For those of you unable to find the reading on recursion over
natural numbers and for completeness, here is the
;;; Procedure: ;;; count-from ;;; Parameters: ;;; lower, a natural number ;;; upper, a natural number ;;; Purpose: ;;; Construct a list of the natural numbers from lower to upper, ;;; inclusive, in ascending order. ;;; Produces: ;;; ls, a list ;;; Preconditions: ;;; lower <= upper ;;; Both lower and upper are numbers, exact, integers, and non-negative. ;;; Postconditions: ;;; The length of ls is upper - lower + 1. ;;; Every natural number between lower and upper, inclusive, appears ;;; in the list. ;;; Every value in the list with a successor is smaller than its ;;; successor. ;;; For every natural number k less than or equal to the length of ;;; ls, the element in position k of ls is lower + k. (define count-from (lambda (lower upper) (if (= lower upper) (list upper) (cons lower (count-from (+ lower 1) upper)))))
Tuesday, 12 September 2000 [Samuel A. Rebelsky]
Friday, 15 September 2000 [Samuel A. Rebelsky]
Sunday, 18 February 2001 [Samuel A. Rebelsky]
Friday, 27 September 2002 [Samuel A. Rebelsky]
Thursday, 30 January 2003 [Samuel A. Rebelsky]
Friday, 31 January 2003 [Samuel A. Rebelsky]
For those who finish early.
Wednesday, 28 January 2004 [Samuel A. Rebelsky]
Friday, 30 January 2004 [Samuel A. Rebelsky]
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